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Proof step 4 from proof sorted-shift-one (previous step)

m + 1 < n0, m + 1 < # A, A[m + 1, A[m]][n1 + 1] < A[m + 1, A[m]][n1], n1 + 1 < n0, n1 + 1 < # A, n ≤ m, n0 ≤ # A, n ≤ n1, ∀ n1. n ≤ n1 ∧ n1 + 1 < n0 → ¬ A[n1 + 1] < A[n1] ⊦ n1 + 1 = m + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + 1 < # A, m + 1 < n0, n1 + 1 < # A, n1 + 1 < n0, A[n1 + 1] < A[m + 1, A[m]][n1], n ≤ n1, n0 ≤ # A, n ≤ m, ∀ n1. n ≤ n1 ∧ n1 + 1 < n0 → ¬ A[n1 + 1] < A[n1] ⊦ n1 = m

Rule argument

Additional information

Used simplifier rules: m ≠ n ⊦ ar[m, a][n] = ar[n]; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)